Difference between revisions of "2019 AMC 12B Problems/Problem 4"
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Dividing both sides by <math>n!</math> gives | Dividing both sides by <math>n!</math> gives | ||
| − | <cmath>(n+1)+(n+2)(n+1)=440 \Rightarrow n^2+4n-437=0 \Rightarrow (n-19)(n+23).</cmath> | + | <cmath>(n+1)+(n+2)(n+1)=440 \Rightarrow n^2+4n-437=0 \Rightarrow (n-19)(n+23)=0.</cmath> |
Since <math>n</math> is positive, <math>n=19</math>. The answer is <math>1+9=10\Rightarrow \boxed{C}.</math> | Since <math>n</math> is positive, <math>n=19</math>. The answer is <math>1+9=10\Rightarrow \boxed{C}.</math> | ||
Revision as of 12:31, 14 February 2019
Problem
A positive integer 
satisfies the equation 
. What is the sum of the digits of ?
Solution
Dividing both sides by 
gives
![\[(n+1)+(n+2)(n+1)=440 \Rightarrow n^2+4n-437=0 \Rightarrow (n-19)(n+23)=0.\]](http://latex.artofproblemsolving.com/9/5/a/95acebc265a9f6c6779a1c3e16f867517e875b6a.png)
Since is positive, 
. The answer is 
See Also
| 2019 AMC 12B (Problems • Answer Key • Resources) | |
| Preceded by Problem 3 |
Followed by Problem 5 |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
| All AMC 12 Problems and Solutions | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
